Problem: A $33$-gon $P_1$ is drawn in the Cartesian plane.  The sum of the $x$-coordinates of the $33$ vertices equals $99$.  The midpoints of the sides of $P_1$ form a second $33$-gon, $P_2$.  Finally, the midpoints of the sides of $P_2$ form a third $33$-gon, $P_3$.  Find the sum of the $x$-coordinates of the vertices of $P_3$.
Answer: Let the $x$-coordinates of the vertices of $P_1$ be $x_1,x_2,\ldots,x_{33}$.  Then, by the midpoint formula, the $x$-coordinates of the vertices of $P_2$ are $\frac{x_1+x_2}2,\frac{x_2+x_3}2,\ldots,\frac{x_{33}+x_1}2 $.  The sum of these equals $\frac{2x_1+2x_2+\cdots +2x_{33}}2=x_1+x_2+\cdots+x_{33}$.  Similarly, the sum of the $x$-coordinates of the vertices of $P_3$ equals the sum of the $x$-coordinates of the vertices of  $P_2$.  Thus the desired answer is $\boxed{99}$.